我已经能够使用围绕给定中心点的纬度和经度值在地图上绘制椭圆。虽然我在地图上看到了一个形状,但是我得到一个椭圆而不是圆形,我认为它与指定的距离不匹配。我打算用它来显示那个圆圈内的物体(一旦我能够正确显示圆圈,这将在以后完成,这就是为什么我需要一个圆而不是椭圆的原因,因为它应该是完美的圆形)。
我正在使用Bing Maps API。我希望从通过参数传入的中心的给定英里(距离)中绘制圆圈,名为miles的参数中的另一个变量只是保持1D的双倍值。我认为问题与我的数学计算方式有关。有没有人知道如何改进这些代码来更好地计算我的里程数。
private void drawPoly(SearchLocation center, Double miles)
{
//amount of vertex
double vertexCount = 100D;
//used by the api to carried out searches
List<SearchLocation> vertices = new List<SearchLocation>();
double v = 0;
double radians = Math.PI / 180D;
double radiansPerDegree = Math.PI / 180D;
double degreePerVertex = 360D / vertexCount;
double radiansPerVertex = degreePerVertex * radiansPerDegree;
var centerOfMap = center;
const double degLatMiles = 68.68637156368D;
double degLonMiles = Math.Cos(center.Latitude.Value) * (68.68637156368D);
double milesLat = (miles * degLatMiles) / 3600;
double milesLon = (miles * degLonMiles) / 3600;
for (v = 0; v < vertexCount; v++)
{
radians = v * radiansPerVertex;
//adds the miles from the center point and draws a circle
double centrLat = center.Latitude.Value + (milesLat * Math.Sin(radians));
double centrLon = center.Longitude.Value + (milesLon * Math.Cos(radians));
vertices.Add(new SearchLocation() { Latitude = centrLat, Longitude = centrLon });
}
答案 0 :(得分:1)
好的,我误解了你的问题。这应该有效:
/// <summary>
/// Calculates the end-point from a given source at a given range (meters) and bearing (degrees).
/// This methods uses simple geometry equations to calculate the end-point.
/// </summary>
/// <param name="source">Point of origin</param>
/// <param name="range">Range in meters</param>
/// <param name="bearing">Bearing in degrees</param>
/// <returns>End-point from the source given the desired range and bearing.</returns>
public static PointLatLng CalculateDerivedPosition(PointLatLng source, double range, double bearing)
{
double latA = source.Lat * DEGREES_TO_RADIANS;
double lonA = source.Lng * DEGREES_TO_RADIANS;
double angularDistance = range / EARTH_RADIUS_M;
double trueCourse = bearing * DEGREES_TO_RADIANS;
double lat = Math.Asin(
Math.Sin(latA) * Math.Cos(angularDistance) +
Math.Cos(latA) * Math.Sin(angularDistance) * Math.Cos(trueCourse));
double dlon = Math.Atan2(
Math.Sin(trueCourse) * Math.Sin(angularDistance) * Math.Cos(latA),
Math.Cos(angularDistance) - Math.Sin(latA) * Math.Sin(lat));
double lon = ((lonA + dlon + Math.PI) % (Math.PI * 2)) - Math.PI;
return new PointLatLng(
lat / DEGREES_TO_RADIANS,
lon / DEGREES_TO_RADIANS);
}
Juste将您的中心作为来源:
for (int i = 0; i < 360; i++)
{
vertices.Add(CalculateDerivedPosition(center, circleRadius, i));
}
答案 1 :(得分:0)
为防止某些纬度上的省略号,我使用以下代码:
// Function to draw circle on map:
private void DrawCircle(BasicGeoposition CenterPosition, int Radius)
{
Color FillColor = Colors.Purple;
Color StrokeColor = Colors.Red;
FillColor.A = 80;
StrokeColor.A = 80;
Circle = new MapPolygon
{
StrokeThickness = 2,
FillColor = FillColor,
StrokeColor = StrokeColor,
Path = new Geopath(Functions.CalculateCircle(CenterPosition, Radius))
};
mpBingMaps.MapElements.Add(Circle);
}
// Constants and helper functions:
const double earthRadius = 6371000D;
const double Circumference = 2D * Math.PI * earthRadius;
public static List<BasicGeoposition> CalculateCircle(BasicGeoposition Position, double Radius)
{
List<BasicGeoposition> GeoPositions = new List<BasicGeoposition>();
for (int i = 0; i <= 360; i++)
{
double Bearing = ToRad(i);
double CircumferenceLatitudeCorrected = 2D * Math.PI * Math.Cos(ToRad(Position.Latitude)) * earthRadius;
double lat1 = Circumference / 360D * Position.Latitude;
double lon1 = CircumferenceLatitudeCorrected / 360D * Position.Longitude;
double lat2 = lat1 + Math.Sin(Bearing) * Radius;
double lon2 = lon1 + Math.Cos(Bearing) * Radius;
BasicGeoposition NewBasicPosition = new BasicGeoposition();
NewBasicPosition.Latitude = lat2 / (Circumference / 360D);
NewBasicPosition.Longitude = lon2 / (CircumferenceLatitudeCorrected / 360D);
GeoPositions.Add(NewBasicPosition);
}
return GeoPositions;
}
private static double ToRad(double degrees)
{
return degrees * (Math.PI / 180D);
}
此代码适用于小于半英里的小半径。